Answer:
The mass of the ice block is equal to 70.15 kg
Explanation:
The data for this exercise are as follows:
F=90 N
insignificant friction force
x=13 m
t=4.5 s
m=?
applying the equation of rectilinear motion we have:
x = xo + vot + at^2/2
where xo = initial distance =0
vo=initial velocity = 0
a is the acceleration
therefore the equation is:
x = at^2/2
Clearing a:
a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2
we use Newton's second law to calculate the mass of the ice block:
F=ma
m=F/a = 90/1.283=70.15 kg
Some examples of stable system are:
1) functions of sine
2) DC
3) signum
4) unit step
5) cosine.
Happy Studying! ^^
It can be described as a constant variation
